What can be said about the values mod $n$ of the Fibonacci polynomial $F_{n}(x)$ ?
A little numerical evidence suggests that
If $p$ is prime, then $F_{n}(k) \equiv -1,0,1 \bmod p$ for all integers $k$.
I'd love to see a proof.
If this is true, I'd like to learn a characterisation of which primes have $F_{p}(k) \equiv 0\bmod p$ for some $k$. This is true for $p=5,13,17,29,37,41,53,\dots$.
Suppose that there are solutions $\alpha, \beta$ to $x^2 - kx - 1 = 0$ in $\mathbb F_p$. Then (working in $\mathbb F_p$) $F_p(k)$ is given by $\frac{\alpha^p - \beta^p}{\alpha - \beta}$ by the usual solution to the linear recurrence. By Fermat's little theorem, $\alpha^p = \alpha$ and $\beta^p = \beta$, so this yields $1$.
When there are no solutions to the quadratic in $\mathbb F_p$, we can work over $\mathbb F_{p^2}$ instead, by adjoining a root $\alpha$ (and getting a second root $\beta = k-\alpha$ for free). The formula $\frac{\alpha^p - \beta^p}{\alpha - \beta}$ is still valid, but $x \mapsto x^p$ is the Frobenius endomorphism, which swaps $\alpha$ and $\beta$, so $\frac{\alpha^p - \beta^p}{\alpha - \beta} = \frac{\beta-\alpha}{\alpha-\beta} = -1$.
The remaining case occurs when $x^2 - kx - 1 = 0$ has a double root $\alpha$ in $\mathbb F_p$, in which case $F_p(k)$ is given by $p \cdot \alpha^{p-1} = 0$. (This form of the solution is correct because $n \cdot \alpha^{n-1}$ is a candidate solution in the case of a double root, and we can check that it's correct for $n=0$ and $n=1$.)
When can a double root occur? When the discriminant $k^2+4$ is $0$ (again, modulo $p$). This can happen only if $-4$ is a quadratic residue modulo $p$; equivalently, if $-1$ is a quadratic residue. We know that this happens only if $p \equiv 1 \pmod 4$.
The Fibonacci polynomial $F_{p}(x)$ factors mod $p$ explicitly as follows: $$ F_{p}(x) = (x^2+4)^{\frac{p-1}{2}} $$
Therefore,
If $k^2+4=0$, then $F_p(k)=0$. This happens iff $p \equiv 1 \bmod 4$.
If $k^2+4\ne0$, then $F_p(k)=(x^2+4)^{\frac{p-1}{2}}=\pm 1$, because of Euler's criterion.